Practice Problems for Indefinite Integration

Integration of $ \displaystyle x^n$

$\displaystyle \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\displaystyle \int{{{{x}^{n}}\ dx=\frac{{{{x}^{{n+1}}}}}{{n+1}}+C}},n\ne -1 } \\ \hline \end{array}\end{array}$

➊       Evaluate each of the following indefinite integrals.

(a) $ \displaystyle \int{{\left( {5{{x}^{3}}-10{{x}^{2}}+4} \right)}}\ dx$

(b) $ \displaystyle \int{{\left( {{{x}^{8}}+\frac{1}{{{{x}^{8}}}}} \right)}\ }dx$

(c) $ \displaystyle \int{{\left( {3\sqrt[4]{{{{x}^{3}}}}-\frac{2}{{{{x}^{5}}}}+\frac{1}{{4\sqrt{x}}}} \right)}\ }dx$

(d) $ \displaystyle \int{{\left( {x-2} \right)(x+2)\ }}dx$

(e) $ \displaystyle \int{{2x(x+3)\ }}dx$

(f) $ \displaystyle \int{{\left( {\frac{{4{{x}^{3}}-2{{x}^{2}}+3x}}{{2x}}} \right)\ }}dx$

(g) $ \displaystyle \int{{\left( {\frac{{2x-3}}{{\sqrt[3]{x}}}} \right)\ }}dx$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \displaystyle \int{{(5{{x}^{3}}-10{{x}^{2}}+4)}}~dx\\\\\ \ \ \ =\ \ \displaystyle \int{{5{{x}^{3}}}}~dx-\displaystyle \int{{10{{x}^{2}}\ }}dx+\displaystyle \int{4}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{5}{4}{{x}^{4}}-\displaystyle \frac{{10}}{3}{{x}^{3}}+4x+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(b)}\ \ \ \ \displaystyle \int{{\left( {{{x}^{8}}+\displaystyle \frac{1}{{{{x}^{8}}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{{{x}^{8}}}}~dx+\displaystyle \int{{\displaystyle \frac{1}{{{{x}^{8}}}}\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{{{x}^{8}}}}~dx+\displaystyle \int{{{{x}^{{-8}}}\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{9}{{x}^{9}}-\displaystyle \frac{1}{7}{{x}^{{-7}}}+C\\\\\ \ \ \ =\ \ \displaystyle \frac{{{{x}^{9}}}}{9}-\displaystyle \frac{1}{{7{{x}^{7}}}}+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(c)}\ \ \ \ \displaystyle \int{{\left( {3\sqrt[4]{{{{x}^{3}}}}-\displaystyle \frac{2}{{{{x}^{5}}}}+\displaystyle {1}{{4\sqrt{x}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{3\sqrt[4]{{{{x}^{3}}}}}}~dx-\displaystyle \int{{\displaystyle \frac{2}{{{{x}^{5}}}}\ }}dx+\displaystyle \int{{\displaystyle \frac{1}{{4\sqrt{x}}}\ }}dx\\\\\ \ \ \ =\ \ 3\displaystyle \int{{{{x}^{{\frac{3}{4}}}}}}~dx-2\displaystyle \int{{{{x}^{{-5}}}\ }}dx+\displaystyle \frac{1}{4}\displaystyle \int{{{{x}^{{-\frac{1}{2}}}}}}~dx\\\\\ \ \ \ =\ \ \displaystyle \frac{{12}}{7}{{x}^{{\frac{7}{4}}}}+\displaystyle \frac{1}{2}{{x}^{{-4}}}+\displaystyle \frac{1}{2}{{x}^{{\frac{1}{2}}}}+C\\\\\ \ \ \ =\ \ \displaystyle \frac{{12{{x}^{{\frac{7}{4}}}}}}{7}+\displaystyle \frac{1}{{2{{x}^{4}}}}+\displaystyle \frac{{\sqrt{x}}}{2}+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(d)}\ \ \ \ \displaystyle \int{{\left( {x-2} \right)\left( {x+2} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {{{x}^{2}}-4} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{{{x}^{2}}\ }}dx-\displaystyle \int{{4\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \frac{1}{3}{{x}^{3}}-4x+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(e)}\ \ \ \ \displaystyle \int{{2x\left( {x+3} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {2{{x}^{2}}-6x} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{2{{x}^{2}}\ }}dx-\displaystyle \int{{6x\ }}dx\\\\\ \ \ \ =\ \ 2\displaystyle \int{{{{x}^{2}}\ }}dx-6\displaystyle \int{{x\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{3}{{x}^{3}}-3{{x}^{2}}+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(f)}\ \ \ \ \displaystyle \int{{\left( {\displaystyle \frac{{4{{x}^{3}}-2{{x}^{2}}+3x}}{{2x}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {\displaystyle \frac{{4{{x}^{3}}}}{{2x}}-\displaystyle \frac{{2{{x}^{2}}}}{{2x}}+\displaystyle \frac{{3x}}{{2x}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {2{{x}^{2}}-x+\displaystyle \frac{3}{2}} \right)\ }}dx\\\\\ \ \ \ =\ \ 2\displaystyle \int{{{{x}^{2}}\ }}dx-\displaystyle \int{{x\ }}dx+\displaystyle \frac{3}{2}\displaystyle \int{{dx}}\\\\\ \ \ \ =\ \ \displaystyle \frac{2}{3}{{x}^{3}}-\displaystyle \frac{1}{2}{{x}^{2}}+\displaystyle \frac{3}{2}x+C\end{array}$ $ \displaystyle \begin{array}{l}\text{(g)}\ \ \ \ \displaystyle \int{{\left( {\displaystyle \frac{{2x-3}}{{\sqrt[3]{x}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ \displaystyle \int{{\left( {2{{x}^{{ \frac{2}{3}}}}-3{{x}^{{-\frac{1}{3}}}}} \right)\ }}dx\\\\\ \ \ \ =\ \ 2\displaystyle \int{{{{x}^{{\frac{2}{3}}}}\ }}dx-3\displaystyle \int{{{{x}^{{-\frac{1}{3}}}}}}\ dx\\\\\ \ \ \ =\ \ \displaystyle \frac{6}{5}{{x}^{{\frac{5}{3}}}}-\displaystyle \frac{9}{2}{{x}^{{\frac{2}{3}}}}+C\end{array}$

➋       Given that $ \displaystyle \frac{{dy}}{{dx}}=3{{x}^{2}}-4x-5$ and $ \displaystyle y=-4$ when $ \displaystyle x=-2,$ find the value of $ \displaystyle y$ when $ \displaystyle x=1.$

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=3{{x}^{2}}-4x-5\\\\\therefore \ \ \ \ dy=\left( {3{{x}^{2}}-4x-5} \right)dx\\\\\therefore \ \ \ \ y=\displaystyle \int{{\left( {3{{x}^{2}}-4x-5} \right)}}dx\\\\\ \ \ \ \ \ y={{x}^{3}}-2{{x}^{2}}-5x+C\\\\\ \ \ \ \ \ \text{When}\ x=-2,\ y=-4\\\\\therefore \ \ \ \ {{\left( {-2} \right)}^{3}}-2{{\left( {-2} \right)}^{2}}-5\left( {-2} \right)+C=-4\\\\\therefore \ \ \ \ C=2\\\\\therefore \ \ \ \ y={{x}^{3}}-2{{x}^{2}}-5x+2\\\\\therefore \ \ \ \ \text{When}\ x=1,\ y=-4\end{array}$

➌       If $ \displaystyle {g}'(x)= 3x^2-2x-3$ and $ \displaystyle g(-1)=4,$ factorize g(x) completely.

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$ \displaystyle \begin{array}{l}\ \ \ \ {g}'(x)=3{{x}^{2}}-2x-3\\\\\therefore \ \ g(x)=\displaystyle \int{{{g}'(x)}}\ dx\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ = \displaystyle \int{{\left( {3{{x}^{2}}-2x-3} \right)}}\ dx\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ = {{x}^{3}}-{{x}^{2}}-3x+C\\\\\ \ \ \ \text{Since}\ g(-1)=4,\\\\\ \ \ \ -1-1+3+C=4\\\\\therefore \ \ C=3\\\\\therefore \ \ g(x)={{x}^{3}}-{{x}^{2}}-3x+3\\\\\ \ \ \ g(1)=1-1-3+3=0\end{array}$ $ \displaystyle \therefore \ \ x-1$ is a factor of $ \displaystyle g(x).$ $\displaystyle \begin{matrix}\begin{array}{r} \left. 1 \right) \\ ~ \end{array} & \underline { \begin{array}{rrrr} 1&-1&-3&3\\ ~~~~&1&0&-3 \end{array} } \\ ~ & \begin{array}{rrrr} \ \ { 1 }& \ \ \ { 0 }& { -3 }&\ \ \color{red}{ 0 } \end{array}\end{matrix}$ $ \displaystyle \begin{array}{l}\therefore \ \ g(x)=(x-1)({{x}^{2}}-3)\\\\\therefore \ \ g(x)=(x-1)(x-\sqrt{3})(x+\sqrt{3})\end{array}$

➍       Given that the curve which has gradient $ \displaystyle \frac{dy}{dx}=2x^2 + 7x$ at any point on the curve passes through the origin, determine the equation of the curve.

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$ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{{dy}}{{dx}}=2{{x}^{2}}+7x\\\\\therefore \ \ dy=\left( {2{{x}^{2}}+7x} \right)dx\\\\\therefore \ \ y=\displaystyle \int{{\left( {2{{x}^{2}}+7x} \right)}}\ dx\\\\\therefore \ \ y=\displaystyle \frac{2}{3}{{x}^{3}}+\displaystyle \frac{7}{2}{{x}^{2}}+C\end{array}$ Since the curve passes through the origin, when $ \displaystyle x=0$ and $ \displaystyle y=0$. $ \displaystyle \begin{array}{l}\therefore \ \ 0=\displaystyle \frac{2}{3}\left( 0 \right)+\displaystyle \frac{7}{2}\left( 0 \right)+C\\\\\therefore \ \ C=0\\\\\therefore \ \ y=\displaystyle \frac{2}{3}{{x}^{3}}+\displaystyle \frac{7}{2}{{x}^{2}}\end{array}$

➎       The rate of change of $ \displaystyle A$ with respect to $ \displaystyle r$ is given by $ \displaystyle \frac{{dA}}{{dr}}=6r+1$. If $ \displaystyle A = 3$ when $ \displaystyle r = 1$, find $ \displaystyle A$ in terms of $ \displaystyle r$.

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$ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{dA}}{{dr}}=6r+1\\\\\therefore \ dA=\left( {6r+1} \right)dr\\\\\therefore \ A=\displaystyle \int{{\left( {6r+1} \right)}}\ dr\\\\\therefore \ A=3{{r}^{2}}+r+C\end{array}$ When $ \displaystyle r=1$ and $ \displaystyle A=3$. $ \displaystyle \begin{array}{l}\therefore \ 3=3{{\left( 1 \right)}^{2}}+\left( 1 \right)+C\\\\\therefore \ C=-1\\\\\therefore \ A=3{{r}^{2}}+r-1\end{array}$

➏       If the curve which has gradient $ \displaystyle \frac{dy}{dx}=kx-1$ at any point on the curve cuts the $ \displaystyle x$-axis at $ \displaystyle x=-3$ and $ \displaystyle x=4$, determine the value of $ \displaystyle k$ and hence find the equation of the curve.

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$ \displaystyle \begin{array}{l}\ \ \ \ \ \displaystyle \frac{{dy}}{{dx}}=kx-1\\\\\therefore \ \ \ dy=\left( {kx-1} \right)dx\\\\\therefore \ \ \ y=\displaystyle \int{{\left( {kx-1} \right)}}\ dx\\\\\ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{k}{2}{{x}^{2}}-x+C\end{array}$ Since the curve cuts the $ \displaystyle x$-axis at $ \displaystyle x=-3$ and $ \displaystyle x=4$ $ \displaystyle \begin{array}{l}\ \ \ \ \ \displaystyle \frac{k}{2}{{\left( {-3} \right)}^{2}}-\left( {-3} \right)+C=0\\\\\therefore \ \ \ 9k+6+2C=0\ ---(1)\\\\\ \ \ \ \ \displaystyle \frac{k}{2}{{\left( 4 \right)}^{2}}-\left( 4 \right)+C=0\\\\\therefore \ \ \ 16k-8+2C=0\ ---(2)\\\\\ \ \ \ \ \text{By}\ (2)-(1),\\\\\ \ \ \ \ 7k-14=0\\\\\therefore \ \ \ k=2\\\\\therefore \ \ \ 9(2)+6+2C=0\ \\\\\therefore \ \ \ C=-12\\\\\therefore \ \ \ y={{x}^{2}}-x-12\ \ \end{array}$

➐       If $ \displaystyle f''\left( x \right) = 15\sqrt x + 5{x^3} + 6$, $ \displaystyle f\left( 1 \right) = - \frac{5}{4}$ and $ \displaystyle f\left( 4 \right) = 404$, find the expression for $ \displaystyle f(x)$.

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$ \displaystyle \begin{array}{l}\ \ \ \ {f}''\left( x \right)=15\sqrt{x}+5{{x}^{3}}+6\\\\\therefore \ \ {f}'\left( x \right)=\displaystyle \int{{{f}''\left( x \right)}}\ dx\\\\\therefore \ \ {f}'\left( x \right)=\displaystyle \int{{\left( {15\sqrt{x}+5{{x}^{3}}+6} \right)}}\ dx\\\\\therefore \ \ {f}'\left( x \right)=10{{x}^{{\displaystyle \frac{3}{2}}}}+\displaystyle \frac{5}{4}{{x}^{4}}+6x+c\\\\\therefore \ \ f\left( x \right)=\displaystyle \int{{{f}'\left( x \right)}}\ dx\\\\\therefore \ \ f\left( x \right)=\displaystyle \int{{\left( {10{{x}^{{ \frac{3}{2}}}}+\displaystyle \frac{5}{4}{{x}^{4}}+6x+c} \right)}}\ dx\\\\\therefore \ \ f\left( x \right)=4{{x}^{{ \frac{5}{2}}}}+\displaystyle \frac{1}{4}{{x}^{5}}+3{{x}^{2}}+cx+d\\\\\ \ \ \ f(1)=-\displaystyle \frac{5}{4}\ \ \ \left[ {\text{given}} \right]\\\\\therefore \ \ 4+\displaystyle \frac{1}{4}+3+c+d=-\displaystyle \frac{5}{4}\\\\\therefore \ \ c+d=-\displaystyle \frac{{17}}{2}\ \ ---(1)\\\\\ \ \ \ f(4)=404\ \ \ \left[ {\text{given}} \right]\\\\\therefore \ \ 128+231+48+4c+d=404\\\\\therefore \ \ 4c+d=-28\ \ ---(2)\\\\\therefore \ \ c=-\displaystyle \frac{{13}}{2}\ \operatorname{and}\ d=-2\\\\\therefore \ \ f\left( x \right)=4{{x}^{{\frac{5}{2}}}}+\displaystyle \frac{1}{4}{{x}^{5}}+3{{x}^{2}}-\displaystyle \frac{{13}}{2}x-2\end{array}$

➑       Given that a ball moves along a grove with the velocity $ \displaystyle v=\frac{{ds}}{{dt}}=32t-2$ ft/s and at the time $ \displaystyle t=\frac{1}{2}\ \text{s}$ the ball is $ \displaystyle 4\ \text{ft}$ away from the starting point. Express the position of the ball as a function of time.

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$ \displaystyle \begin{array}{l}\ \ \ \ v=\displaystyle \frac{{ds}}{{dt}}=32t-2\\\\\therefore \ \ ds=\left( {32t-2} \right)dt\\\\\therefore \ \ s=\displaystyle \int{{\left( {32t-2} \right)}}\ dt\\\\\therefore \ \ s=16{{t}^{2}}-2t+C\\\\\ \ \ \ \text{When}\ t=\displaystyle \frac{1}{2},\ s=4\\\\\therefore \ \ 16{{\left( {\displaystyle \frac{1}{2}} \right)}^{2}}-2\left( {\displaystyle \frac{1}{2}} \right)+C=4\\\\\therefore \ \ C=1\\\\\therefore \ \ s=16{{t}^{2}}-2t+1\end{array}$

Integration of $ \displaystyle (ax + b)^n$

$\displaystyle \begin{array}{l}\begin{array}{|r||l|l|l|c|c|c|c|c|c|} \hline {\displaystyle \int{{{{{\left( {ax+b} \right)}}^{n}}}}\ dx=\displaystyle \frac{{{{{\left( {ax+b} \right)}}^{{n+1}}}}}{{a\left( {n+1} \right)}}+C,} \\ {\text{where}\ n\ne -1,\ a\ne 0 \ \ \ \ \ \ \ \ \ }\\ \hline \end{array}\end{array}$

➒       Integrate each of the following with respect to $ \displaystyle x$.

(a) $ \displaystyle (2x+1)^6$

(b) $ \displaystyle \frac{2}{{\sqrt{{3x-2}}}}$

(c) $ \displaystyle (5x-11)^{10}$

(d) $ \displaystyle {{\left( {\frac{3}{{4x-1}}} \right)}^{3}}$

(e) $ \displaystyle \frac{2}{{5{{{\left( {3x-1} \right)}}^{4}}}}$

(f) $ \displaystyle \sqrt{{{{{\left( {5x+2} \right)}}^{3}}}}$

(g) $ \displaystyle \frac{5}{{2\sqrt{{{{{\left( {3x-1} \right)}}^{7}}}}}}$

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$ \displaystyle \begin{array}{l}\text{(a)}\ \ \ \ \ \displaystyle \int{{{{{(2x+1)}}^{6}}}}\ dx\ \\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{7\times 2}}{{(2x+1)}^{7}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{14}}{{(2x+1)}^{7}}+C\\\\\\\\\text{(b)}\ \ \ \ \ \displaystyle \int{{\displaystyle \frac{2}{{\sqrt{{3x-2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ 2\displaystyle \int{{{{{\left( {3x-2} \right)}}^{{- \frac{1}{2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ 2\times \displaystyle \frac{{{{{\left( {3x-2} \right)}}^{{\frac{1}{2}}}}}}{{\displaystyle \frac{1}{2}\times 3}}\ +C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{4}{3}\sqrt{{3x-2}}+C\\\\\\\\\text{(c)}\ \ \ \ \ \displaystyle \int{{{{{(5x-11)}}^{{10}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{11\times 5}}{{(5x-11)}^{{10}}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{55}}{{(5x-11)}^{{10}}}+C\\\\\\\\\text{(d)}\ \ \ \ \ \displaystyle \int{{{{{\left( {\displaystyle \frac{3}{{4x-1}}} \right)}}^{3}}}}\ dx\\\\\ \ \ \ \ =\ \ 27\displaystyle \int{{{{{\left( {4x-1} \right)}}^{{-3}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{{27}}{{-2\times 4}}{{\left( {4x-1} \right)}^{{-2}}}+C\\\\\ \ \ \ \ =\ \ -\displaystyle \frac{{27}}{{8{{{\left( {4x-1} \right)}}^{2}}}}+C\\\\\\\\\text{(e)}\ \ \ \ \ \displaystyle \int{{\displaystyle \frac{2}{{5{{{\left( {3x-1} \right)}}^{4}}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{2}{5}\displaystyle \int{{{{{\left( {3x-1} \right)}}^{{-4}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{2}{{5(-3)(3)}}{{\left( {3x-1} \right)}^{{-3}}}+C\\\\\ \ \ \ \ =\ \ -\displaystyle \frac{2}{{45{{{\left( {3x-1} \right)}}^{3}}}}+C\\\\\\\\\text{(f)}\ \ \ \ \ \displaystyle \int{{\sqrt{{{{{\left( {5x+2} \right)}}^{3}}}}}}\ dx\\\\\ \ \ \ \ =\ \displaystyle \int{{{{{\left( {5x+1} \right)}}^{{\frac{3}{2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{1}{{\displaystyle \frac{5}{2}\times 5}}{{\left( {5x+1} \right)}^{{\displaystyle \frac{5}{2}}}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{2}{{25}}{{\left( {5x+1} \right)}^{{\frac{5}{2}}}}+C\\\\\\\\\text{(g)}\ \ \ \ \ \displaystyle \int{{\displaystyle \frac{5}{{2\sqrt{{{{{\left( {3x-1} \right)}}^{7}}}}}}}}\ dx\\\\\ \ \ \ \ =\ \displaystyle \frac{5}{2}\displaystyle \int{{{{{\left( {3x-1} \right)}}^{{-\frac{7}{2}}}}}}\ dx\\\\\ \ \ \ \ =\ \ \displaystyle \frac{5}{2}{{\left( {3x-1} \right)}^{{- \frac{5}{2}}}}+C\\\\\ \ \ \ \ =\ \ \displaystyle \frac{5}{{2\left( {-\displaystyle \frac{5}{2}} \right)3}}{{\left( {3x-1} \right)}^{{- \frac{5}{2}}}}+C\\\\\ \ \ \ \ =\ \ -\displaystyle \frac{1}{{3{{{\left( {3x-1} \right)}}^{{ \frac{5}{2}}}}}}+C\end{array}$

➓       Find the equation of the curve which cuts the $ \displaystyle y$-axis at $ \displaystyle (0, 4)$ and for which $ \displaystyle \frac{{dy}}{{dx}}={{(2x+1)}^{3}}$.

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$ \displaystyle \begin{array}{l}\ \ \ \displaystyle \frac{{dy}}{{dx}}={{(2x+1)}^{3}}\\\\\therefore \ \ dy={{(2x+1)}^{3}}dx\\\\\therefore \ \ y=\displaystyle \int{{{{{(2x+1)}}^{3}}}}\ dx\\\\\therefore \ \ y=\displaystyle \frac{1}{{4\times 2}}{{(2x+1)}^{4}}+C\\\\\therefore \ \ y=\displaystyle \frac{1}{8}{{(2x+1)}^{4}}+C\end{array}$ Since the curve passes through $ \displaystyle (0, 4)$, $ \displaystyle \begin{array}{l}\ \ \ \ \displaystyle \frac{1}{8}{{(2(0)-1)}^{4}}+C=4\\\\\therefore \ \ \ C=\displaystyle \frac{{31}}{8}\\\\\therefore \ \ y=\displaystyle \frac{1}{8}{{(2x+1)}^{4}}+\displaystyle \frac{{31}}{8}\\\\\therefore \ \ y=\displaystyle \frac{1}{8}\left[ {{{{(2x+1)}}^{4}}+31} \right]\end{array}$

Sequences and Series

Sequence



ေအာက္မွာေပးထားတဲ့ ကိန္းတန္းေလးကို ေလ့လာၾကည့္ရေအာင္။



1,4,9,16,25,...



အခုကိန္းတန္းမွာပါ၀င္တဲ့ ကိန္းလံုးေတြဟာ ေရးခ်င္သလို ေရးထားျခင္း (random) မဟုတ္ပါဘူး။ ကိန္းလံုးေတြ ေျပာင္းလဲမႈမွာ စနစ္တစ္ခု (တနည္းေျပာရရင္ function တစ္ခု) အရ ေျပာင္းလဲသြားတာပါ။ ဒါကို functional notation နဲ႔ေျပာရမယ္ဆိုရင္...



f(1) = 1 = 1²

f(2) = 4 = 2²

f(3) = 9 = 3²

f(4) = 16 = 4²

f(5) = 25 = 5² လို႔ဆိုႏိုင္တာေပါ။့



ဒါကိုၾကည့္ျခင္းအားျဖင့္ function ရဲ့ Domain ဟာ {1,2,3,4,5,...n}=the set of natural numbers ဆိုရပါမယ္။ ဒါဆိုရင္ အထက္ပါေျပာင္းလဲမႈကို ၾကည့္ရံုနဲ႔ function ရဲ့ general formula ကို အလြယ္တကူ ေျပာႏိုင္ပါၿပီ။

f(n) = n² ေပါ့...။



ဒါေၾကာင့္ sequence ဆိုတာဟာ special function လို႔ဆိုႏိုင္ၿပီး သူရဲ့ domain ကေတာ့ အၿမဲတမ္း သဘာ၀ကိန္း မ်ား ပါ၀င္ေသာအစု (the set of natural numbers) ျဖစ္ပါတယ္။ image ေတြကိုေတာ့ ဒီေနရာမွာ terms လို႔ ေျပာင္းလဲေခၚပါမယ္။ အေခၚအေ၀ၚ ေျပာင္းလဲမႈနဲ႔အတူ အသံုးျပဳမယ့္ symbols ေတြကိုလည္း ေျပာင္းလည္း သတ္မွတ္ပါတယ္။ အထက္မွာေပးထားတဲ့ ကိန္းလံုးေတြကို အခုလိုေခၚေ၀ၚ သတ္မွတ္ပါမယ္။

first term	=	u1	=	1
second term	=	u2	=	4
third term	=	u3	=	9
fourth term	=	u4	=	16
fifth term	=	u5	=	25
- - - - - - - -	- -	- -	- -	- -
nth term	=	un	=	n2

ဒီေနရာမွာ n^th term ကို general term သို႔မဟုတ္ general formula လို႔ဆိုႏိုင္ပါတယ္။ ဒါေၾကာင့္ sequence ကို အခုလို definition ဖြင့္ဆိုႏိုင္ပါတယ္။



Sequence

A sequence is a function whose domain is either the set of all or part of natural numbers. The values (images) of function are called terms .



Example 1

Find the first four terms of the sequence defined by u_n = 3n - 5.



Solution

u_n = 3n - 5

u₁ = 3(1) - 5 =-2

u₂ = 3(2) - 5 = 1

u₃ = 3(3) - 5 = 4

u₄ = 3(4) - 5 = 7

Therefore the fist four terms are -2, 1, 4, 7.



Exercises

Find the first four terms of the sequence defined by

(a) un = 2n + 3

(b) un = 3n2 - 2

(c) un = 4n2+ 3n - 5

Example2

Which term of the sequence defined by u_n = 4n - 23 is 25?



Solution

u_n = 4n - 23

Let the n^th term be 25.

Therefore u_n = 25

4n - 23 = 25

4n = 48

n = 12

Therefore the 12^th term is 25.

Miscellaneous Problems

Find the remainder when 3x³ - x² + 7x + 5 is divided by 3x + 2.

Given that 2x³ - x² - 2x + 3 = (Ax + B)(x - 1)(x + 2) + C(x - 1) + D, find the values of A, B, C and D. Hence or otherwise, deduce the remainder when

2x³ - x² - 2x + 3 is divided by x² + x - 2.

Solve the equation 2x³ + x² - 19x = 6, giving your answers to two decimal places where necessary.

The expression x³ + px² + qx + 6 has the same remainder when divided by

x + 1 and 2 - x. Given that the remainder when the expression is divided by x + 3 is -60, find the value of p and of q.

By remainder theorem,

remainder = f(-2/3)

= 3(-2/3)³ - (-2/3)² + 7(-2/3) + 5

= -1

Let x = 1

2 - 1 - 2 + 3 = D

D = 2

Let x = -2

-16 - 4 + 4 + 3 = -3C + 2

3C = 15

C = 5

Let x = 0

3 = B (-1)(2) -5 + 2

-2B = 6

B = -3

Sub any value other than 1, 2 & 0 into x

A = 2

2x³ - x² - 2x + 3 ≡ (Ax + B)(x - 1)(x + 2) + C(x - 1) + D

Since x² + x - 2 = (x - 1)(x + 2), (note the similarity?)

thus, remainder = C(x - 1) + D

= 5(x - 1) + 2

= 5x - 3

2x³ + x² - 19x = 6

let f(x) = 2x³ + x² - 19x - 6

(x - 3) f(3) = 0

(x - 3) is a factor of f(x)

let f(x) = (x - 3)(Ax² + Bx + C)

A = 2

C = 2

let x = 1

2 + 1 - 19 - 6 = -2(2 + B + 2)

-22 = -8 - 2B

B = 7

f(x) = (x - 3)(2x² + 7x + 2)

2x² + 7x + 2 = 0

x =

= -3.19 or -0.31 (3 s.f.)

x = -3.19 or -0.31 or 3

let f(x) = x³ + px² + qx + 6

by remainder theorem, f(-1) = f(2)

-1 + p - q + 6 = 8 + 4p + 2q + 6

p - q - 5 = 4p + 2q + 14

3p + 3q = 9 --------(1)

by remainder theorem, f(-3) = -60

-27 + 9p - 3q + 6 = -60

3q = 39 + 9p --------(2)

sub (2) into (1):

3p + 39 + 9p = -9

12p = -48

p = -4

q = 1

Factor Theorem အသံုးခ် ပုစာၦမ်ား

Example 2

Given that x² + x - 6 is a factor of 2x⁴ + x³ - ax + bx + a + b - 1, find the value of a and b.

x² + x - 6 = (x + 3)(x + 2)

Let f(x) = 2x⁴ + x³ - ax² + bx + a + b - 1

f(-3) = 2(-3)⁴ + (-3)³ - a(-3)² - 3b + a + b - 1 = 0

134 - 8a - 2b = 0

4a + b = 67 --------(1)

f(2) = 2(2)⁴ + 2³ - a(2)² + 2b + a + b - 1 = 0

39 - 3a + 3b = 0

a - b = 13 --------(2)

(1) + (2) : 5a = 80

a = 16

when a = 16, b = 3

Example 3

Use the factor theorem to find the value of k for which (a + 2b) where a does not = 0 and b does not = 0, is a factor of a⁴ + 32b⁴ + a³b(k + 3).

Solution



Let f(a) = a⁴ + 32b⁴ + a³b(k + 3)

f(-2b) = (-2b)⁴ + 32b⁴ + (-2b)³b(k + 3) = 0

48b⁴ - 8b⁴(k+ 3) = 0

8b⁴[6 - (k + 3)] = 0

8b⁴(3 - k) = 0

Since b does not = 0, 3 - k = 0

k = 3

Example 4

Determine the value of k for which x + 2 is a factor of (x + 1)⁷ + (2x + k)³.

Solution



Let f(x) = (x + 1) + (2x + k)

f(-2) = (-2 + 1) + (-4 + k) = 0

(k - 4) - 1 = 0

(k - 4) = 1

k - 4 = 1

k = 5

Example 5

Given that f(x) = x³ - x² - ax - b, where a and b are constants, has the factor x - 3 but has a remainder of 13x - 11 when divided by x + 4. Calculate the values of a and b.

Hence (i) factorise f(x) completely. (ii) Solve the equation 27x³ - 3ax = ax² + b.

Solution



f(3) = 0

3³ - 3² - 3a - b = 0

3a + b = 18 --------(1)

f(-4) = (-4)³ - (-4)² - a(-4) - b = 13(-4) - 11

4a - b = 17 --------(2)

sub (1) into (2): a = 5, b = 3

(i) Factorising.

f(x) = (x - 3)(x² + 2x + 1)

= (x - 3)(x + 1)²

(ii) 27x³ - 9x² - 3ax - b = 0

(3x)³ - (3x)² - a(3x) - b = 0

Let y = 3x

y³ - y² - ay - b = 0

(y - 3)(y + 1)² = 0 [from (1)]

y = 3, y = -1

x = 1, x = -1/3

Example 6

Show that the expression x³ + (k-2) x² + (k-7)x - 4 has a factor x+1 for all values of k .If the expression also has a factor x+2, find the value of k and the third factor.



Solution

Let f(x) = x³ + (k-2) x² + (k-7)x - 4

f(-1) =(-1)³ + (k-2)(-1)² + (k-7)(-1) - 4 = -1+ k - 2 - k + 7 - 4 = 0

Therefore (x+1) is a factor of f(x).

x + 2 is a factor of f(x). -------(given)

Therefore f(-2)=0

(-2)³ + (k-2)(-2)² + (k-7)(-2) - 4 = 0

-8 + 4k -8 - 2k +14 - 4 = 0

2k = 6

k = 3

Therefore f(x) = x³ + x² - 4x - 4

Therefore (x + 1) (x + 2) = x² + 3x + 2 is also a factor of f(x).

Therefore the third factor of f(x) is x - 2.



Example 7

Given that kx³ + 2x² + 2x + 3 is a factor of kx³ - 2x + 9, have a common factor, what are the possible values of k?



Solution

Let f(x) = >kx³ + 2x² + 2x + 3 and g(x) = kx³ - 2x + 9.

Let (x - c) be a common factor of both f(x) and g(x).Therefore ...

f(c) = 0

kc³ + 2c² + 2c + 3 = 0 ---------(1)



g(c) = 0

kc³ - 2c + 9= 0 ---------(2)



Subtracting equation(2) from (1),we have

2c² + 4c - 6 = 0

c² + 2c - 3 = 0

(c + 3) (c - 1) =0

c = -3 or c = 1



Substituting c = -3 or c = 1 in equation (2), we get

k = 5/9 when c = -3 and

k = -7 when c = 1

Remainder Theorem ကိုအသံုးခ်တြက္ေသာ ပုစာၦမ်ား(3)

Example 10

The remainder when x⁴ + 3x² - 2x + 2 is divided by x + a is the square of the remainder when x² - 3 is divided by x + a, Calculate the possible values of a.
x⁴ + 3x² - 2x + 2 ကို x + a ႏွင့္စားေသာအခါ ရတဲ့ remainder သည္ x² - 3 ကို x + a ႏွင့္ စားေသာအခါ ရတဲ့ remainder၏ ႏွစ္ထပ္ႏွင့္ ညီလွ်င္ a ၏ တန္ဖိုးမ်ားကို ရွာပါ။

Let f(x) = x⁴ + 3x² - 2x + 2 and g(x) = x² - 3
By the problem,
f(-a) = [g(-a)]²
(-a)⁴ + 3(-a)² - 2(-a) + 2 =[(-a)² - 3]²
a⁴ + 3a² + 2a + 2 = a⁴ - 6a² + 9
9a² + 2a - 7 = 0
(a + 1) (9a - 7) = 0
a = -1 (or) a=7/9

Example 11

The expression ax³ - x² + bx - 1 leaves the remainder of -33 and 77 when divided by x + 2 and x - 3 respectively. Find the values of a and b and the remainder when divided by x - 2.
ax³ - x² + bx - 1 ကို x + 2 နဲ႔စားတဲ့အခါ -33 ရၿပီး x - 3 နဲ႔စားတဲ့အခါ 77 ရပါတယ္။ a ႏွင့္ b ကိုရွာပါ။ x - 3 ႏွင့္စားလို႔ရတဲ့ remainder ကိုလည္းရွာေပးပါ။

Let f(x) = ax³ - x² + bx - 1.
By the problem,
f(-2) = -33
a(-2)³ - (-2)² + b(-2) - 1 = -33
4a + b = -33 -------------(1)

f(3) = 77
a(3)³ - (3)² + b(3) - 1 = 77
9a + b = 29 -------------(2)

equation (2) - equation (1),we get
5a = 15 and a = 3

Substitute a = 3 in equation (1)
4(3) + b = 14
b = 2

Remainder Theorem ကိုအသံုးခ်တြက္ေသာ ပုစာၦမ်ား(2)

Example 8

Given that the expression x³ - ax² + bx + c leaves the same remainder when divided by x + 1 or x - 2 , find a in terms of b.
ေပးထားေသာ ကိန္းတန္း x³ - ax² + bx + c ကို x + 1 ႏွင့္ x - 2 တို႔ျဖင့္စားေသအခါ တူညီေသာ အၾကြင္းရ၏။ a ၏ တန္ဖိုးကို b ပါ၀င္ေသာ ကိန္းတန္းျဖင့္ ေဖၚျပပါ။

Let f(x) = x³ - ax² + bx + c.
By the problem,
f(-1) = f(2)
(-1)³ - a(-1)² + b(-1) + c = (2)³ - a(2)² + b(2) + c
-1 - a -b = 8 - 4a + 2b
a = b + 3

Example 9

Given that the remainder when x³ - x² + ax is divided by x + a where a>0, is twice the remainder when it is divided by x - 2a, find the value of a.
x³ - x² + ax ကို x + a ႏွင့္စား၍ ရေသာ အၾကြင္းသည္ x - 2a ႏွင့္စား၍ ရေသာ အၾကြင္း၏ ၂ ဆျဖစ္လွ်င္ a ၏ တန္ဖိုးကို ရွာပါ။ (မွတ္ခ်က္ a>0)

Let

By the problem,

Since a>0,

Remainder Theorem ကိုအသံုးခ်တြက္ေသာ ပုစာၦမ်ား(1)

Example 6

The polynomial x³ + ax² + bx - 3 leaves a remainder of 27 when divided by x - 2 and a remainder of 3 when divided by x + 1.Calculate the remainder when the polynomial is divided by x - 1

x³ + ax² + bx - 3 ကို x - 2 နဲ႔စားရင္ remainder 27 ရတယ္။ x + 1 နဲ႔ စားရင္ေတာ့ remainder 3 ရတယ္။ x - 1 နဲ႔ စားလို႔ရတဲ့ remainder ကိုရွာေပးပါ။

ရွင္းလင္းခ်က္။ ။ ေပးထားတဲ့ polynomial ကို f(x)လို႔ထားမယ္။ f(x) ကို x - 1 နဲ႔စားရင္ remainder f(1) ေပါ့။ ဒါေပမယ့္ unknown constant a နဲ႔ b ကိုသိရမယ္။ အဲဒါကို ေပးခ်က္ႏွစ္ခ်က္ကေန အရင္ရွာမယ္။ a နဲ႔ b ကို သိရင္ ဆက္ရွာ လို႔ ရၿပီေပါ့။

Let f(x) = x³ + ax² + bx - 3, by the problem

f(2) = 27

2³ + a(2²) + 2b - 3 = 27

2a + b = 11 -------------(1)

f(-1) = 3

(-1)³ + a(-1)² - b - 3 = 27

a - b = 31 ----------------(2)

equation (1) + equation (2) => 3a = 42 and a = 14

Substitute a = 14 in equation (2),

14 - b = 31

b = -17

Therefore f(x) = x³ + 14x² - 17x - 3

f(x) ÷ (x - 1) => remainder = f(1)

f(1) = 1³ + 14(1²) - 17(1) - 3 = - 5

Example 7

The expression x³ - 7x + 6 and x³ - x² - 4x +24 have the same remainder when divided by x + p. Find the possible values of p.
ကိန္းတန္းႏွစ္ခု ေပးထားပါတယ္။ ဒါေၾကာင့္ ပထမကိန္းတန္းကို f(x) လို႔ထားမယ္၊ ဒုတိယကိန္းတန္းကို g(x) လို႔ ထားမယ္။ ႏွစ္ခုလံုးကို x + p နဲ႔ စားတဲ့အခါ remainder တူတယ္။ p ရဲ့တန္ဖိုး ရွာေပးရမယ္။

Let f(x) = x³ - 7x + 6 and g(x) = x³ - x² - 4x +24
By the problem, f(-p) = g(-p)
p³ - 7p + 6 = p³ - p² - 4p +24
p² - 3p - 18 = 0
(p - 6) (p + 3) = 0
p = 6 or p = -3